A balanced lever has two weights on it, the first with mass #3 kg # and the second with mass #4 kg#. If the first weight is # 8 m# from the fulcrum, how far is the second weight from the fulcrum?

1 Answer
Jan 23, 2017

Answer:

6 m away from the fulcrum, opposite from the first weight.

Explanation:

For the lever to be "balanced", there must be no net torque on the system, or

#tau_"net"=0#

or

#Sigma tau = 0#

Assuming the fulcrum is at the center of mass of the lever (or the lever is massless), the torques created by each of the weights must be equal and opposite:

#tau_1 + tau_2 = 0#
#tau_1 = - tau_2#

Torque is the force perpendicular to a point of rotation multiplied by the distance from that pivot point. Assuming the lever to be horizontal, the torque equation becomes:

#F_1xxr_1 = -(F_2xxr_2)#
#m_1gxxr_1=-(m_2gxxr_2)#

Solving for #r_2# gives:

#r_2=((m_1cancel(g))r_1)/(-m_2cancel(g))=-m_1/m_2r_1#

Applying the given information, we get:

#r_2=-(3"kg")/(4"kg")(8"m")=-6"m"#

The fact that #r_2# is a negative value tells us that it is on the opposite side of the fulcrum from the first weight.

The second weight must be positioned 6 m away from the fulcrum on the other side of the fulcrum.